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Energy and momentum operators in QM

by McLaren Rulez
Tags: energy, momentum, operators
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McLaren Rulez
#1
Jun18-11, 04:09 AM
P: 261
Hi,

I understand that we use [itex]i\hbar\partial/\partial t[/itex] and [itex]-i\hbar\nabla[/itex] for the energy and momentum operator but I would like to know how this identification is made.

I can see that it works for a wave of the form [itex]e^{i(kx-\omega t)}[/itex] and using the relation [itex]E=\hbar\omega[/itex] and the relation [itex]p=h/\lambda[/itex].

But what about everything else? How did physicists come to the conclusion that the general energy and momentum operator are of the form mentioned above?

Thank you very much.
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dextercioby
#2
Jun18-11, 04:38 AM
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There's no energy operator, rather the Hamilton operator. And the way these operators <look> in quantum mechanics (what they are represented through) depends a lot on the system under analysis. The so-called <canonical commutation relations> set forth at the end of the 1920's, within the properly defined mathematical environment (the theory of separable Hilbert spaces), lead inevitably to a representation of the momentum operator in coordinate space as [itex] -i\hbar \nabla [/itex] (Schroedinger 1926, Stone 1930 and von Neumann 1931).

As for the explicit form of the Hamilton operator, this is largely dependent on the quantization scheme and on whether the quantum system has a classical counterpart whose Hamiltonian formalism is fully known.
McLaren Rulez
#3
Jun18-11, 06:08 AM
P: 261
Dextercioby, thank you for the quick reply.

So ultimately, how did Schrodinger write down his equation? I thought the motivation behind it was something along the lines of my original post.

I understand how [itex][x\ p]=i\hbar[/itex] leads to the momentum operator but what about the Hamiltonian operator? Since my QM text starts off with the Schrodinger eqaution, I would like to know a little about how Schrodinger himself came up with it.

Thank you.

P.S. Could you also tell me what is the difference between the energy operator and the Hamiltonian operator? I was under the mistaken assumption that they were the same thing.

Lapidus
#4
Jun18-11, 06:11 AM
P: 283
Energy and momentum operators in QM

Quote Quote by McLaren Rulez View Post
Hi,

I understand that we use [itex]i\hbar\partial/\partial t[/itex] and [itex]-i\hbar\nabla[/itex] for the energy and momentum operator but I would like to know how this identification is made.

I can see that it works for a wave of the form [itex]e^{i(kx-\omega t)}[/itex] and using the relation [itex]E=\hbar\omega[/itex] and the relation [itex]p=h/\lambda[/itex].

But what about everything else? How did physicists come to the conclusion that the general energy and momentum operator are of the form mentioned above?

Thank you very much.
Momentum is the generator of the space translation, energy is the generator of the time translation of a state. From there you can see why they have the form they have.

Look for example here
http://www.hep.upenn.edu/~rreece/doc...esentation.pdf

or google translation generator operator quantum....or something like that
dextercioby
#5
Jun18-11, 06:32 AM
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Quote Quote by McLaren Rulez View Post
Dextercioby, thank you for the quick reply.

So ultimately, how did Schrodinger write down his equation? I thought the motivation behind it was something along the lines of my original post.
Essentially he built on L. de Broglie's idea of matter waves for particles (p=h \nu) and used the theory of electromagnetic waves as a starting point in deriving his equation. He ended up with his famous equation whose predictions he successfully tested against a model for the hydrogen atom (obtained the countable nature of negative energy levels).

His 4 original articles in German in the <Annalen der Physik> have been translated into English (a book, google for <Collected papers on wave mechanics>) and he also published a lucid summary of his work in the American journal <The Physical Review> in 1926.

Quote Quote by McLaren Rulez View Post
I understand how [itex][x\ p]=i\hbar[/itex] leads to the momentum operator but what about the Hamiltonian operator? Since my QM text starts off with the Schrodinger eqaution, I would like to know a little about how Schrodinger himself came up with it.

Thank you.

P.S. Could you also tell me what is the difference between the energy operator and the Hamiltonian operator? I was under the mistaken assumption that they were the same thing.
There's no energy operator, but merely the Hamiltonian operator which is somehow related to the Hamiltonian of the classical theory (where the latter exists).

On a superficial level, all classical observables of the Hamiltonian formalism are <promoted> to linear operators on a linear space and above all one postulates the Schroedinger equation for the <wavefunction> (2nd order partial differential equation).

For a beginner's level, it's the usual approach, since the knowledge of mathematics is kept to a minimum.

P.S. From his 4 articles one sees that he was aware of the <Dreimaennerarbeit> of Heisenberg, Jordan and Born (1925) which founded <matrix mechanics>. In 1926 he went on to prove the equivalence of matrix and wave mechanics in a separate paper.
McLaren Rulez
#6
Jun18-11, 03:23 PM
P: 261
Thank you. I will take a look at the original papers when I get the chance as well as the translation operator link.

One final question: If the Hamiltonian operator is related to the classical Hamiltonian, then it must have something to do with the total energy, right? Is it just a matter of calling it the Hamiltonian instead of energy operator or is there something deeper that I am missing here?

Thank you once again.
dextercioby
#7
Jun19-11, 05:04 AM
Sci Advisor
HW Helper
P: 11,926
Quote Quote by McLaren Rulez View Post
Thank you. I will take a look at the original papers when I get the chance as well as the translation operator link.

One final question: If the Hamiltonian operator is related to the classical Hamiltonian, then it must have something to do with the total energy, right? Is it just a matter of calling it the Hamiltonian instead of energy operator or is there something deeper that I am missing here?

Thank you once again.
Yes, the convention is to name it as Hamilton operator.
McLaren Rulez
#8
Jun19-11, 06:22 AM
P: 261
Thank you very much dextercioby.


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